# 物理代写|电动力学代写electromagnetism代考|Clausius-Mossotti Equation

#### Doug I. Jones

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## 物理代写|电动力学代写electromagnetism代考|Clausius-Mossotti Equation

One further point needs discussion. In our atomic model for the dielectric constant in Section 5.2, we used the following equation for the electronic motion, (5.16):
$$m \frac{d^2}{d t^2} \mathbf{r}=-m \omega_0^2 \mathbf{r}-m \gamma \frac{d}{d t} \mathbf{r}+e \mathbf{E}$$
Here the driving force has been taken to be $e \mathbf{E}$ where the macroscopic field is $\mathbf{E}(\mathbf{r}, t)=\overline{\mathbf{e}(\mathbf{r}, t)}$. The same assumption for the driving field was made for the alignment of polar molecules in Section 5.4. This is incorrect since e includes the field of the atom (or molecule) itself, the effects of which are already represented in the harmonic restoring force. The correct driving field is the field acting on the electron due to all the other atoms (we use the word atom to stand, generically, for either atom or molecule),
$$\mathbf{E}{\text {driving }}=\overline{\mathbf{e}{\text {driving }}}=\overline{\mathbf{e}-\mathbf{e}{\text {atom }}}=\mathbf{E}-\overline{\mathbf{e}{\text {atom }}},$$
where now the overbar represents a spatial average over a volume $V$ which contains exactly one atom, that average volume per atom being the inverse of the density,
$$n=\frac{1}{V}$$
[In (5.78) we assume that no significant contribution is produced by neighboring atoms, so that $\mathbf{E}{\text {driving }}$ does not differ appreciably from $\mathbf{e}{\text {driving }}$, and also that the average over the atom is already sufficiently representative of a macroscopic average that the field $\mathbf{E}$ can be introduced. What we are doing should not be expected to apply for a strongly polar liquid or solid where the forces produced by neighboring molecules could be the dominant effect.] The field due to the atom in which the electron is located is
$$\mathbf{e}{\mathrm{atom}}=-\nabla \sum_a \frac{e_a}{\left|\mathbf{r}-\mathbf{r}_a\right|}=\sum_a e_a \nabla_a \frac{1}{\left|\mathbf{r}-\mathbf{r}_a\right|},$$ where the summation extends over all charges in the atom. We can calculate $\overline{\mathbf{e}{\text {atom }}}$ by averaging over a sphere (of radius $a$ and volume $V=4 \pi a^3 / 3$ ) which is large enough to include the atom; the negative of the average field can be written as
$$-\overline{\mathbf{e}_{\mathrm{atom}}}=\sum_a\left(-\nabla_a\right) e_a \frac{1}{V} \int_V \frac{(d \mathbf{r})}{\left|\mathbf{r}-\mathbf{r}_a\right|} .$$

## 物理代写|电动力学代写electromagnetism代考|Canonical Equations of Motion in Electromagnetic Fields

Since magnetic effects are more subtle than electric ones, it is helpful to first develop the formalism describing a charge moving in the presence of electric and magnetic fields. We start with the equation of motion for such a charge,
$$m \frac{d}{d t} \mathbf{v}=e\left(\mathbf{E}+\frac{\mathbf{v}}{c} \times \mathbf{B}\right)$$
Although, unlike an electric field, a magnetic field does no work on a charge, there is a magnetic term in the energy, because the act of turning on a magnetic field produces an electric field, according to Faraday’s law, (1.64). To see this, it is convenient to recast (6.1) so that potentials appear instead of field strengths. Since the magnetic field has zero divergence,
$$\boldsymbol{\nabla} \cdot \mathbf{B}=0,$$
we may introduce a vector potential, $\mathbf{A}$, such that
$$\mathbf{B}=\boldsymbol{\nabla} \times \mathbf{A} .$$
When we substitute (6.3) into Faraday’s law
$$-\frac{1}{c} \frac{\partial}{\partial t} \mathbf{B}=\nabla \times \mathbf{E}$$
we observe that the quantity $\mathbf{E}+\frac{1}{c} \frac{\partial}{\partial t} \mathbf{A}$ is irrotational,
$$\boldsymbol{\nabla} \times\left(\mathbf{E}+\frac{1}{c} \frac{\partial}{\partial t} \mathbf{A}\right)=\mathbf{0}$$
which means that a corresponding scalar potential $\phi$ exists,
$$\mathbf{E}+\frac{1}{c} \frac{\partial}{\partial t} \mathbf{A}=-\nabla \phi$$
This is the generalization of (1.13) to incorporate time dependence. In terms of the potentials $\mathbf{A}$ and $\phi$, the equation of motion (6.1) reads
$$m \frac{d}{d t} \mathbf{v}=-e \boldsymbol{\nabla} \phi-\frac{e}{c} \frac{\partial}{\partial t} \mathbf{A}+\frac{e}{c} \mathbf{v} \times(\boldsymbol{\nabla} \times \mathbf{A})$$
which can be rewritten as
$$m \frac{d}{d t} \mathbf{v}=-\frac{e}{c} \frac{d}{d t} \mathbf{A}-e \boldsymbol{\nabla}\left(\phi-\frac{1}{c} \mathbf{v} \cdot \mathbf{A}\right),$$
since ( $\boldsymbol{\nabla}$ operates only on $\mathbf{A}$ ),
$$\mathbf{v} \times(\boldsymbol{\nabla} \times \mathbf{A})=\boldsymbol{\nabla}(\mathbf{v} \cdot \mathbf{A})-(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}$$
and
$$\frac{\partial}{\partial t} \mathbf{A}(\mathbf{r}, t)+(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}(\mathbf{r}, t)=\frac{d}{d t} \mathbf{A}(\mathbf{r}(t), t) .$$
This suggests that we define the canonical momentum of the particle by
$$\mathbf{p}=m \mathbf{v}+\frac{e}{c} \mathbf{A}$$
which supplements the kinetic momentum $m \mathbf{v}$ by the potential momentum $\frac{\epsilon}{c} \mathbf{A}$. In terms of $\mathbf{p}$ the equation of motion reads
$$\frac{d}{d t} \mathbf{p}=-\nabla\left(e \phi-\frac{e}{c} \mathbf{v} \cdot \mathbf{A}\right)$$

# 电动力学代考

## 物理代写|电动力学代写electromagnetism代考|Clausius-Mossotti Equation

$$m \frac{d^2}{d t^2} \mathbf{r}=-m \omega_0^2 \mathbf{r}-m \gamma \frac{d}{d t} \mathbf{r}+e \mathbf{E}$$

$$\mathbf{E}{\text {driving }}=\overline{\mathbf{e}{\text {driving }}}=\overline{\mathbf{e}-\mathbf{e}{\text {atom }}}=\mathbf{E}-\overline{\mathbf{e}{\text {atom }}},$$

$$n=\frac{1}{V}$$
[在式(5.78)中，我们假定邻近的原子没有产生显著的贡献，因此$\mathbf{E}{\text {driving }}$与$\mathbf{e}{\text {driving }}$没有明显的差别，而且原子上的平均值已经足以代表宏观平均值，因此可以引入$\mathbf{E}$场。我们所做的不应该适用于强极性液体或固体，在那里邻近分子产生的力可能是主要作用。由于电子所在的原子所产生的场为
$$\mathbf{e}{\mathrm{atom}}=-\nabla \sum_a \frac{e_a}{\left|\mathbf{r}-\mathbf{r}a\right|}=\sum_a e_a \nabla_a \frac{1}{\left|\mathbf{r}-\mathbf{r}_a\right|},$$原子中所有电荷的总和。我们可以计算$\overline{\mathbf{e}{\text {atom }}}$通过平均一个球(半径$a$和体积$V=4 \pi a^3 / 3$)，这是大到足以包括原子;平均场的负值可以写成 $$-\overline{\mathbf{e}{\mathrm{atom}}}=\sum_a\left(-\nabla_a\right) e_a \frac{1}{V} \int_V \frac{(d \mathbf{r})}{\left|\mathbf{r}-\mathbf{r}_a\right|} .$$

## 物理代写|电动力学代写electromagnetism代考|Canonical Equations of Motion in Electromagnetic Fields

$$m \frac{d}{d t} \mathbf{v}=e\left(\mathbf{E}+\frac{\mathbf{v}}{c} \times \mathbf{B}\right)$$

$$\boldsymbol{\nabla} \cdot \mathbf{B}=0,$$

$$\mathbf{B}=\boldsymbol{\nabla} \times \mathbf{A} .$$

$$-\frac{1}{c} \frac{\partial}{\partial t} \mathbf{B}=\nabla \times \mathbf{E}$$

$$\boldsymbol{\nabla} \times\left(\mathbf{E}+\frac{1}{c} \frac{\partial}{\partial t} \mathbf{A}\right)=\mathbf{0}$$

$$\mathbf{E}+\frac{1}{c} \frac{\partial}{\partial t} \mathbf{A}=-\nabla \phi$$

$$m \frac{d}{d t} \mathbf{v}=-e \boldsymbol{\nabla} \phi-\frac{e}{c} \frac{\partial}{\partial t} \mathbf{A}+\frac{e}{c} \mathbf{v} \times(\boldsymbol{\nabla} \times \mathbf{A})$$

$$m \frac{d}{d t} \mathbf{v}=-\frac{e}{c} \frac{d}{d t} \mathbf{A}-e \boldsymbol{\nabla}\left(\phi-\frac{1}{c} \mathbf{v} \cdot \mathbf{A}\right),$$

$$\mathbf{v} \times(\boldsymbol{\nabla} \times \mathbf{A})=\boldsymbol{\nabla}(\mathbf{v} \cdot \mathbf{A})-(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}$$

$$\frac{\partial}{\partial t} \mathbf{A}(\mathbf{r}, t)+(\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}(\mathbf{r}, t)=\frac{d}{d t} \mathbf{A}(\mathbf{r}(t), t) .$$

$$\mathbf{p}=m \mathbf{v}+\frac{e}{c} \mathbf{A}$$

$$\frac{d}{d t} \mathbf{p}=-\nabla\left(e \phi-\frac{e}{c} \mathbf{v} \cdot \mathbf{A}\right)$$

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